Question: $f(x) = \begin{cases} 3\sqrt{x } & \text{for} ~~~~x\gt{4} \\ 2x-8& \text{for} ~~~~ x \leq4\end{cases}$ Evaluate the definite integral. $\int^9_{3}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $12$ (Choice B) B $13$ (Choice C) C $26$ (Choice D) D $37$
Explanation: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^9_{3}f(x)\,dx$ $= \int^9_{4}f(x)\,dx + \int^{4}_{3}f(x)\,dx~~~~~~$ [Why did we split at 4?] $= \int^9_{4}3\sqrt{x}\,dx + \int^4_{3}(2x-8)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^9_{4}3\sqrt{x}\,dx &=2x^\frac32\Bigg|^9_{{4}} \\\\ &= \left[2 ( 9)^\frac32 \right] - \left[2 ({4})^\frac32\right] \\\\ &= \left[54\right] -\left[16 \right] \\\\ &= {38}\end{aligned}$ The second definite integral: $\begin{aligned} \int^4_{3}(2x-8)\,dx &=(x^2-8x)\Bigg|^4_{{3}} \\\\ &= \left[ ( 4)^2 - 8\cdot(4)\right] - \left[( 3)^2 - 8\cdot(3)\right] \\\\ &= \left[-16\right] -\left[-15 \right] \\\\ &= {-1}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^9_{4}3\sqrt{x}\,dx + \int^4_{3}(2x-8)\,dx$ $ = {38} + ({-1})$ $ = 37$ The answer $\int^9_{3}f(x)\,dx = 37$